This is one way to do it. You are just projecting J_tor to a finite
element space and use that for output or further processing. As others
have already noted, the only problem I can see is that computing
solution_max may be inaccurate. This is because in general, if you have
a finite element function
u_h(x) = \sum_i \varphi_i(x) U_i
then
max_{x \in \Omega} u_h(x)
is not the same as
max_i U_i
For (bi-,tri-)linear functions the two are the same but for higher order
finite elements, that is not the case.
you mean that solution is not u_h(x) but U_i. is it right?
Well, both are the solution -- one is a function of x, the other is a vector
of nodal values, but they ultimately represent the same object. U_i is the
vector you get out of the linear solver.
Then, is there other way to get maximum value of solution for higher order
finite elements?
Not easily. To see the problem, think about using quadratic finite elements in
1d. Let's say you have only one cell [0,1], so the shape functions are defined
at x=0, x=1/2, and x=1. Assume you have nodal values 1,0,0, then your solution
vector is
u_h(x) = sum_i U_i \varphi_i(x) = 1 * \varphi_0(x)
= 2(1-x)(1/2-x)
= 1 - 3x + 2x^2
Note that while the minimal value of U_i is zero, the function u_h(x) has a
mimimum at u_h'(x)=-3+4x=0, i.e., at x=3/4 where its value is
u_h(3/4) = 1 - 9/4 + 2*9/16 = -1/8
In other words, even though your nodal values are 1,0,0, the minimum of this
function is -1/8 < 0 !
In general, the location and the value of the minimum depend in complex ways
on the nodal values and can not easily be predicted. The best you can do is to
use a quadrature formula with sufficiently many points and sample the solution
at these points to get upper bounds for the minimum, and lower bounds for the
maximum of the solution function u_h(x).
Best
W.
--
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Wolfgang Bangerth email: [email protected]
www: http://www.math.colostate.edu/~bangerth/
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